Problem: $f(t) = \begin{cases} t^2-5t &, & t\leq -10 \\\\ t+19 &, & -10<t<-2\\\\ \dfrac{t^3}{t+9} &, & t \geq -2\end{cases}$ $f(-10)=$
Solution: The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $t={-10}$. Finding the appropriate assignment rule Since ${-10}\leq-10$, we should use the first assignment rule $t^2-5t$. The answer $f({-10})=({-10})^2-5({-10})=150$ In conclusion, $f(-10)=150$.